"Veritasium" (YT) - "The Big Misconception About Electricity" ?

[rfeecs]

Energy is present in the fields.  You know that there is energy in the electric E field in a capacitor for example.  There is energy in the magnetic B field in an inductor for example.

There is no electric field inside a good conducting wire.  E=0 inside the wire.  If there were an E field, the charge would move until there is no E field.  Similarly there is no changing magnetic field inside the wire.

So the energy is in the fields.  The conductors just set the boundary conditions (zero tangential E-field at the surfaces, etc.)

An example is a waveguide.  The conductor walls just serve to contain the fields.  The energy is in the fields that propagate through the space in between.

The same for the two wires.  They form a transmission line.  A two conductor line will have a TEM mode (Transverse Electro Magnetic) that propagates down to DC.

When the switch is first closed, the two wire transmission line looks just like a resistor equal to the characteristic impedance of the line, maybe about 600 ohms in this case.  So the initial current will flow through the switch, through the resistor and through the light.

Of course the fields can't propagate instantly, so it will take at least 1m/c to move the one meter from the switch to the light.

[rfeecs]

If you don't get the idea of a transmission line, you can think of two wires that are thousands of kilometers long as a big capacitor.

When the switch is thrown, you immediately get current through that capacitor which also goes through the light.

Of course there is also distributed inductance in the wires so that will limit the current.

[bdunham7]

So the energy is in the fields.  The conductors just set the boundary conditions (zero tangential E-field at the surfaces, etc.)

No dispute.  The issue, IMO, is how those fields are propagated.  A conductor, with its extremely low permittivity, transmits a static or low frequency field much more efficiently than space.  If you connect a long pair of wires to a 9-volt battery, the 9 volt field quickly appears across the other end of the cable.  The 9 volt battery doesn't power anything by radiating a field through space, even though in theory it does.  And if you short it with your keys, that also causes a field to be radiated.

An example is a waveguide.  The conductor walls just serve to contain the fields.  The energy is in the fields that propagate through the space in between.

Again, absolutely true--for microwaves in waveguides.  In the case of a 9 volt battery and two wires, after the initial connection the resulting field between the two wires is static. 

A two conductor line will have a TEM mode (Transverse Electro Magnetic) that propagates down to DC.

And unless I'm mistaken, as you approach DC the characteristic impedance approaches infinity, or the leakage conductance of the dielectric if any.

When the switch is first closed, the two wire transmission line looks just like a resistor equal to the characteristic impedance of the line, maybe about 600 ohms in this case.  So the initial current will flow through the switch, through the resistor and through the light.

Again, agreed after rethinking a bit.  However, I think the resistance will appear to be twice the characteristic impedance because the line goes both ways--but that is a minor issue.  Also, rethinking the practical aspects, the current won't increase after the 1 second unless you posit a superconductor for the wires because the DC resistance of such a long cable is going to be higher than the characteristic impedance of the pair. 

[rfeecs]

A two conductor line will have a TEM mode (Transverse Electro Magnetic) that propagates down to DC.

And unless I'm mistaken, as you approach DC the characteristic impedance approaches infinity, or the leakage conductance of the dielectric if any.

I agree that the impedance goes up at very low frequencies.  But here we are talking about the first few nanoseconds after the switch is thrown.  We are dealing with high frequencies.

Clearly the resistance of the wire will screw up the whole thing.  Ultimately the DC resistance will be so high that you won't get much current and the light probably won't go on at all in the end.

You have to assume that the resistance of the wire is very low.  Veritasium states that "the wires have no resistance, otherwise this wouldn't work."

From the transcript of the video:

You have to make some simplifying assumptions
about this circuit,
like the wires have to have no resistance,
otherwise this wouldn't work
and the light bulb has to turn on immediately
when current passes through it.

[bdunham7]

But here we are talking about the first few nanoseconds after the switch is thrown.  We are dealing with high frequencies.

Yes, true enough.  Batteries, light bulbs and .... nanoseconds.  OK!

Veritasium states that "the wires have no resistance, otherwise this wouldn't work."

OK, so it won't work.  And if we posit wires with no resistance, we can make them infinitesimally small, which will make the characteristic impedance arbitrarily high!  And with zero resistance, the eventual current flow will be much higher than the initial bit through the characteristic impedance. 

I'll repeat my initial claim that this is clickbaity gibberish. 

[rfeecs]

I agree the initial quiz question is an unnecessary distraction from presenting the concept that the energy flows in the fields and not the charges moving in the wires.

You seem to agree that energy flows with a propagating wave.  But for DC it's a different matter.

What's interesting to me is something he didn't show on the video:

So after several seconds, everything calms down and there is a constant DC current through the wires.  So since the wires are perfect conductors, there is zero E field in the region of the kilometers long wires (Top and bottom wire are at the same potential).  The only significant E field is in the region near the battery and the light.  So according to the theory, energy is only flowing from the battery to the light in that small one meter long region.  There is no energy flowing in the region of the long wires.

[bdunham7]

energy flowing

Can energy 'flow'?  If so, what does the term mean?  A static e-field doesn't transfer any energy.  The infinite permittivity of the (super)conductor allows the e-field of the be transmitted down the wire without loss by allowing electrons to literally, physically flow.  Which is what he was trying to refute and 'blow our minds' with at the beginning of his video, with a mishmash of half-baked gibberish about how electrons don't actually ever get from a generating plant to our home.  That, of course is literally true in the same way that the actual water in the core of a nuclear reactor doesn't (in any sane design) ever actually flow through the steam turbines and cooling towers.  Obviously there is still flowing water inside the system.

[IanB]

Can energy 'flow'?

In the sense that engineers routinely do energy balances around parts of a system, by considering flows of energy in and out of a control envelope, then yes, energy can 'flow'.

One has to remember that in thermodynamics, energy is simply a concept of 'something' that is, by observation and experiment, conserved.

[bdunham7]

Can energy 'flow'?

In the sense that engineers routinely do energy balances around parts of a system, by considering flows of energy in and out of a control envelope, then yes, energy can 'flow'.

One has to remember that in thermodynamics, energy is simply a concept of 'something' that is, by observation and experiment, conserved.

OK, I'm willing to accept any definition for arguments sake.  So can you draw a diagram of how the energy in this case 'flows' from the battery to the light in the steady-state DC case (with superconductors, apparently) ?

[IanB]

OK, I'm willing to accept any definition for arguments sake.  So can you draw a diagram of how the energy in this case 'flows' from the battery to the light in the steady-state DC case (with superconductors, apparently) ?

I'm not sure what you are asking? We can measure the power consumed by the lamp, by measuring the voltage and current at the lamp terminals, and we can measure the power delivered by the battery, by measuring the current and voltage likewise. We will find by experiment that the lamp power is always less than or equal to the battery power. We can carefully exclude any external sources of power into the system. We will eventually conclude that there is a flow of energy from the battery to the lamp. If there were no flow of energy, the lamp would not light up (which might be the actual experimental result).

We do not need to know how the energy got from the battery to the lamp, what mechanism was employed, or what path it took, in order to conclude that the transfer took place.

[bdunham7]

We do not need to know how the energy got from the battery to the lamp, what mechanism was employed, or what path it took, in order to conclude that the transfer took place.

If that is the case and the mechanism of transfer is unknown, how can you be sure that the battery isn't actually powering a different lamp somewhere else and that the lamp isn't being powered by a different battery?  And before you go switching it on and off, I can posit that I've installed sensors on the remote battery and lamp.

Leaving that silly example aside, the fact remains that even if you are convinced that the power flows from the battery to the lamp, you still haven't determined the mechanism and path of that flow.  So, for the hypothetical example, can you explain the mechanism and path by which the energy flows to the battery? 

You can start by identifying the phenomena by which the energy appears in the lamp--an electric field across its terminals results in electrons flowing through it, they bump into things along the way and dissipate energy inside the filament which heats it up--or something like that.

[rfeecs]

So can you draw a diagram of how the energy in this case 'flows' from the battery to the light in the steady-state DC case (with superconductors, apparently) ?

It's not so easy for my feeble brain to visualize.  The flow is out from the battery and in to the light bulb.

You would have to draw the E-field and the B-field and then the energy flow direction is E cross B.

As you go away from the battery, the E field will drop off so most of the flow is in the region close to the battery and the light.

[T3sl4co1l]

A full field model isn't so easy to visualize, for the reasons covered by the first explanation here:

Edit2: Analysis slides using transmission line models linked by Derek: https://ve42.co/bigcircuit [Would light up at 1/c but would take ~2 sec (for the 1 light second total width case) to reach peak]

Namely, that you're considering the superposition of common and differential mode waves, one of which disperses readily (roughly inverse with distance, proportional to frequency of the components), the other which remains between the lines, at least given ideal enough geometry.

Note this means, at a great enough distance, and given some approximation so as to ignore what CM energy remains at that given distance: the driven line will go up at Vdiff/2 while its partner goes down to -Vdiff/2.  This is obvious enough when you consider the TL as a transformer, and it's just transformer action making a balun.  Except it's not really a transformer, it's wires in space radiating the common mode -- but we can use a transformer plus termination resistances to model the same thing in a compact structure.

When you're doing full fields, you can still do the same decomposition, but if you're thinking about it whole, as \$\vec{E}(\vec{x}, t)\$ or \$\vec{B}(\vec{x}, t)\$ from instant to instant, you will quickly run into trouble because it's a superposition of waves at different velocities, and the solution is tricky.  Recognizing decompositions (superposition of modes) is key to solving problems like this.

We could further complicate matters by noting that, at least where the lines are in proximity to the Earth (and, for expedience, we might simply wrap the lines around the Earth many times, rather than actually launching them straight-line into space ), ground effect, and the effective dielectric constant of air and the ground*, act to slow the CM wave, trapping it towards the surface, causing dispersion, scattering and dissipation.  This applies to whatever part of the DM field interacts with the surface, too.

*In general, we can treat the ground as a material of varying, complex, permittivity (and permeability if applicable).  In the same way that complex impedance (reactance) is conservative rather than dissipative (real resistance), imaginary permittivity is dissipative rather than conservative.  Probably dry soil and rock will be mostly real, while wet soil, water and ocean (and magma, deeper still) will contribute a significant imaginary component.

Tim

[emece67]

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[Kalvin]

Would this be similar situation to two parallel conductors at close proximity from each other, a steady state initial condition, and an em-field generated between the two conductors when the switch is closed?

[Kalvin]

Here is a quick simulation of the experiment using LTSpice, and modeling the long wires as lossless transmission lines. And yes, current & energy will flow "instantly" in the load when the switch is closed, so 1m/c is the correct answer. The simulation also shows that that the load will not get full power when the switch is closed. Using a longer simulation time and leaving the switch into the closed state, it can be observed that the load will finally get the full power after the circuit has reached steady state again.

[RoGeorge]

That's not the same schematic.  The switch must be near battery, not near bulb, like this:



If you place the switch near the battery, like Veritasium did, you'll get a much funnier response.

Either way, LTSpice or any other SPICE based simulators are not physics simulators, they are not aware of the speed of light.  They can only simulate lumped circuits where everything propagates instantaneously.

[Kalvin]

Changed the switch position, and simulation looks similar to me. LTSpice is able to simulate transmission lines pretty well, too.

[T3sl4co1l]

That's not the same schematic.  The switch must be near battery, not near bulb, like this:

Does it really, though?





Consider the loop equivalent here:



If the lamp, battery and switch are close together (close enough that we can consider them a short loop, that size being relative to the risetime of the switch; or simply ignoring nanosecond dynamics), then the two transmission lines act in series as impedances, just wherever, and we can rearrange the circuit without loss of generality.

Note in particular, the battery can be placed anywhere along any line; it's a supernode that has no effect on the dynamics of the system whatsoever.  We do have to be careful that, when constructing the circuit, we allow it to come to equilibrium first; this is trivial in analysis as we can assume equilibrium for all t <= 0 with the switch flipped for t > 0.  (And indeed this is what SPICE will do, when DC .op is performed, i.e. not "set to zero".)

Mind, this makes a crucial assumption, that common mode can be ignored -- in the full-field case with setup as proposed, the current flows into CM cannot be rearranged so easily.  Though in this simple case I don't think it actually makes a difference.

Tim

[Trader]

Dave?

[eti]

Dave?

A lone image floating in space with no context... we aren't psychic.

[BrianHG]

That's not the same schematic.  The switch must be near battery, not near bulb, like this:



If you place the switch near the battery, like Veritasium did, you'll get a much funnier response.

Either way, LTSpice or any other SPICE based simulators are not physics simulators, they are not aware of the speed of light.  They can only simulate lumped circuits where everything propagates instantaneously.

     For any doubters out there,  never mind that the wire loops at each end at the 1/2 light second distance.  Even if the ends were open, the lamp would still receive power in the time it takes light to travel 1 meter from switch & battery to lamp the moment the switch is turned on.  You are looking at a huge transformer or antenna with on side transmit and the other receive.  Without the wires connected at each end, it's just that with a DC power source, the lamp would turn on, then run out of power after a second whereas having the loop shorted on the left and right side means the light would stay on with DC power.  Feeding AC tuned to the wire length means the light would stay on without the ends connected.

[EEVblog]

My replay on the video was:

Correct. But your video kind of implies that the diameter of the copper cable therefore does not matter, if it is "just the field on the outside". So that's not the whole story, Mr Ohm wants a word with you.
Not to mention wave propagation time, the RF engineers would like a word with you as well.
Every engineer knows about electron drift velocity and how slow it is, nothing mysterious there at all. But practical engineering design usually ignores such physics detail for practical reasons.
Does current through a capacitor? In practical electronics design, yep, it does.

[Kleinstein]

One can even calculate how much current would flow initially: the long cables are transmission lines with a characteristic impedance. With 1 m distance (and huge diameter to kepp the resistance low) likely with a charcteristic impedance somewhere in the 100 ohms range (no more than 370 Ohms for the free space).

For the initial time the transmission line act just like resistors of the characteristic impedance, only with some delay (1 second for the link at the end) the end and wire resistance is seen.

[Kalvin]

Here is a simulation for 10 seconds (and the corresponding LTSpice files). It is possible to see how the energy propagates through the long wires [lossless transmission lines] back and forth, and how the system converges gradually to the steady state, and how the load will then receive full power. I increased the load to 200 ohms, so that we would see some reflections.

Note: I had to add .txt file extension to the plot settings file due to forum filename restrictions. Just rename the plot settings file as Veritasium_long_cable_c.plt.